I need a panel mount pot which should be capable from swinging from 950 Ohm to 6kOhm. Can be greater or less so long as it satisfies my range. 24VDC at not much more than 20mA max. Not too fussed about teh accuracy, say about 10%
Just looking for a cheapo pot, any suggestions?

Hi,
I want to build a 1-5V, 4-20mA generator. Plan on using a 250Ohm resistor (R2), but am unsure of the selection of pot (R1) to control the voltage. My calc tells me I need the range as per my first post, hence the question for a suitable pot that will cover that for me?

It sounds like you want an adjustable regulated current source to drive an LED through it's complete range from dark to maximum brightness. If that's what you want, you can use an LM317 regulator with a 62Ω resistor and a 250Ω potentiometer in series (62Ω - 312Ω range) as the control resistance in a constant current configuration.

The resistor and potentiometer can be very low wattage since they carry very little of the circuit current; the regulator handles most of it.

You will need a supply voltage that's a few volts more than your LED Vf at 20mA. That's a 6.5V - 7V supply for a typical white LED, less for other colors and the 24V you have available could drive as many as 5 white LEDs in series.

The pot must be rated for at least 2W for a 5k pot and a 20mA maximum current. Even if it doesn't carry the maximum current at the maximum resistance setting you should still select the pot based upon its current rating from the I^2*R power dissipated in the element.

Hi,
I want to build a 1-5V, 4-20mA generator. Plan on using a 250Ohm resistor (R2), but am unsure of the selection of pot (R1) to control the voltage. My calc tells me I need the range as per my first post, hence the question for a suitable pot that will cover that for me?

Many thanks

Click to expand...

What's the plan? Is this for testing industrial 4-20mA current loop devices?
If you have a device that sources 5V and needs a 4-20mA input signal, you can make a dummy load with a 250Ω resistor and a 1KΩ pot in series, but if you are trying to generate 4-20mA, it's a little more complicated.

What's the plan? Is this for testing industrial 4-20mA current loop devices?
If you have a device that sources 5V and needs a 4-20mA input signal, you can make a dummy load with a 250Ω resistor and a 1KΩ pot in series, but if you are trying to generate 4-20mA, it's a little more complicated.

Click to expand...

I do, I also have an old ATX power supply that i could use too as a source. I was always going for the volt divider as you describe, just with a higher Vin, to accomdate my 24V discretes too, but wasnt vital. might just use the atx power supply. This 1k pot, will it swing from 1K down to near 0? any part No.?

I do, I also have an old ATX power supply that i could use too as a source. I was always going for the volt divider as you describe, just with a higher Vin, to accomdate my 24V discretes too, but wasnt vital. might just use the atx power supply. This 1k pot, will it swing from 1K down to near 0? any part No.?

Many thanks

Click to expand...

The problem is loading. For example, put a 250Ω resistor in series with a 1KΩ pot, apply 5V, and you will observe a 4-20mA readout with your meter in series. Now, you're "generating" 4-20mA, but that's with no load. This is only beneficial for testing displays, and really, not even then, because your display is going to have an input impedance. Slap a load in series with your resistor/pot device and now the whole thing goes out the window. For example, I have 0-120PSI current/air pressure transducer herre on my desk. It uses a voice coil to regulate air pressure; 4mA = 0PSI, and 20mA = 120PSI. I ohm out the coil and read 430Ω. If I tried to use this resistor/pot device to run the transducer up & down between 4 & 20 mA, it wouldn't work. If I turn the pot all the way to 0 with no load I should get 20mA (5V/250Ω = 20mA) but now I have 430 extra ohms in the circuit so all I get is 7.35mA (5V/680Ω = 7.35mA) or about 25PSI when I should get 120PSI.

This is why I say you need a current source, which is a little more complicated. If I want to drive this 430Ω coil to 20mA and get 120PSI out of it, I need to apply 8.6V to it, which is out of the range of your 5V power supply. The current source should be able to apply whatever voltage necessary to drive whatever predetermined current through whatever load you chose to connect it to.

Ok, well my goal is to inject this signal into a DAQ, or PLC. If i understand you correctly, i really need to provide a bigger psu, since at 20mA presume I will drop 5V over my 250Ohm resistor, but this wont be taking into account other loads, i.e. cable, input impedance of the DAQ etc?

So, if i went for say a 12V supply from the same PSU, and used a 3K pot, would this suffice?

So I measured an input impedance of 110kOhm at the AIs on the DAQ.
I have done some new calcs taking into consideration the RLoad this time, and have concluded that I could use a 12V supply and R1 could be off the shelf pot, with a value of 4700KOhm.

Using Vin=12V, R1=4700Ohm, R2=250Ohm, and RL as 110KOhm. Then my Vout = 0.605V

Swinging the pot back to say 300Ohm, and using the other parameters as before, then my Vout = 5.442V

This configuration appears to give me the full range of volt drops drop I require over this 250Ohm resistor with respect to my DAQ input.
Does this sound about right? Or have i done something not quite right?

So I measured an input impedance of 110kOhm at the AI’s on the DAQ.
I have done some new calcs taking into consideration the RLoad this time, and have concluded that I could use a 12V supply and R1 could be off the shelf pot, with a value of 4700KOhm.

Using Vin=12V, R1=4700Ohm, R2=250Ohm, and RL as 110KOhm. Then my Vout = 0.605V

Swinging the pot back to say 300Ohm, and using the other parameters as before, then my Vout = 5.442V

This configuration appears to give me the full range of volt drops drop I require over this 250Ohm resistor with respect to my DAQ input.
Does this sound about right? Or have i done something not quite right?

Thanks again for your/everyone’s support.

Click to expand...

Ok I don't follow; I thought you were trying to generate 4-20mA for an analog current input. your 110KΩ analog input is a voltage input, not current; that's way too high for a current input. So the purpose this whole time was to turn 4-20mA into 1-5V? or was it just to generate a 1-5V signal? I'm assuming now that you just want a 1-5V signal

I haven't seen any schematic so I don't know what R1, R2 is - I get the same numbers when I put the 250ohm resistor connected to gnd, and 4.7kohm pot connected to +12V, and the point where they connect in the middle is going to the 110kohm load, is that correct? do you realize that if you turn your pot all the way to zero, it will put 12V on the Analog input? Is it rated for that?
To be safe, try this: Change your 250Ω resistor to a 6.8KΩ resistor, and then Swap your resistor and and pot; pot on the bottom, in parallel with the analog input (load), and resistor on top. Now do the math, you get 0-4.78V and no risk of damaging your DAQ.

Hi, strantor. I did actually realise after i posted Re the 12V when the pot is at 0 Ohm., which as you suggest isnt desired at all.
I also like what you are suggesting, however, would another way not be to just go back with the 5V supply and use a 2500K ohm pot this time? Or add an additional 250 ohm resistor in series with the other 250ohm to bias the input to the daq at 6 volts when the R pot =0ohms? And still use a 2500K pot (schem attached)
The advertised impedance of the DAQ is actually 144K. Within LabVIEW i can define that input to be measuring current, and can also define that Im using an external 250Ohm shunt resistor.
This is the daq http://sine.ni.com/ds/app/doc/p/id/ds-218/lang/en

would another way not be to just go back with the 5V supply and use a 2500K ohm pot this time?

Click to expand...

Definately, but it doesn't necessarily have to be 2500KΩ, it could be any value (as long as it's not such a low value that it exceeds the power supply capacity), and you don't need an additional resistor. Just use all 3 leads of the pot to make the voltage divider. +5V to one end, wiper to the DAQ input, and 0V to the other end of the pot, and to the daq input. That will automatically give you 0-5V into the DAQ. if you really have your heart set on 1-5V, then add a resistor between the pot and ground. You want to size your resistor to create as close to 20% voltage divider as possible. For example, if you are using a 2.5KΩ pot, use a 560Ω resistor between the pot & ground, as this gives you an 18.3% division, or about .92V when the pot is all the way down.
I haven't taken into account the input impedence in this math, because it is so high that it is negligible. Input impedence for voltage inputs is always high just for this reason, so that you can hook up just about whatever you want to it, and it won't load the circuit. If you were using pot & resistor values of several megohm, it would be different. So try to keep the values reasonable, like 2500ohms, that's good - not too high, not too low. When I was asking abot input impedence before, it was because i thought we were dealing with a current input and not a voltage input.